Is $L^{p_1}(\mu)\cap L^{p_0}(\mu)$ dense in $L^r(\mu)$?

Question

Suppose $0<p_0<r<p_1<\infty$. It is known that $L^{p_1}(\mu)\cap L^{p_0}(\mu)\subset L^r(\mu)$. Is it true that $L^{p_1}(\mu)\cap L^{p_0}(\mu)$ dense in $L^r(\mu)$?

Answer 1

Yes. Simple functions composed of sets with finite measure are dense in $L^r(\mu)$ and are contained in $L^{p_1}(\mu) \cap L^{p_0}(\mu)$.

Depending on your particular measure space (for example, if it is $\mathbb R^n$ with the Lebesgue measure), you may also be able to use continuous compactly supported functions.